[R] nls, how to determine function?

[Katrina Bennett]

Hi R help,

I am trying to determine how nls() generates a function based on the
self-starting SSlogis and what the formula for the function would be.
I've scoured the help site, and other literature to try and figure
this out but I still am unsure if I am correct in what I am coming up
with.


**************************************************************************
dat <- c(75.44855206,NA,NA,NA,82.70745342,82.5335019,88.56617647,80.00128866,94.15418227,86.63987539,93.91052952,74.10612245,86.62289562,90.47961047,NA,NA,82.45320197,72.14371257,NA,71.44104803,72.59742896,68.36363636,NA,NA,61,NA,NA,71.26502909,NA,85.93333333,84.34248284,79.00522193,79.64223058,97.2074017,88.43700548,96.40413877,95.13511869,92.57379057,93.97498475,NA,97.55995131,89.53321146,97.21728545,93.21980198,77.54054054,95.85392575,86.25684723,97.55325624,80.03950617,NA,91.34023128,92.42906574,88.59433962,65.77272727,89.63772455,NA,NA,NA,NA,74.86344239,83.57594937,70.22516556,65.30543319,NA,NA,67.84852294,60.90909091,54.79303797,NA,52.18735363,33.47003155,NA,41.34693878,24.5047043,NA,NA,NA,NA,9.944444444,13.6875,NA,11.90267176,84.14285714,3.781456954,NA,1.432926829,4.26557377,1.823529412,0.444620253,4.711155378,NA,6.320284698,0.581632653,0.144578313,3.666666667,0,0,0,0,0,NA,0.032947462,0,0,10.54545455,0,NA,0.561007958,0.75,NA,0.048780488,0.74137931,NA,2.023339318,0,0,0,NA,NA,0.156950673,NA,0.283769634,32.81818182,NA,NA,0,NA,0,0,0,NA,0.212454212,3.120181406,NA,0.011811024,NA,0,0.120430108,5.928571429,1.75,0.679292929,0.97,NA,0,NA,NA,1,0.38547486,NA,1.460732984,0.007795889,0.05465288,0.004341534)
dat.df.1 <- data.frame(dat)
dat.df.2 <- data.frame(x=x.seq, dat.df=dat.df.1)
fit.dat <-nls(dat ~ SSlogis(x, Asym, xmid,scal), data = dat.df.2,
start =list(Asym=90, xmid = 75, scal = -6))
plot(dat.df.2, axes=FALSE, ann=FALSE, ylim=c(0,100))
lines(dat.df.2$x[complete.cases(dat.df.2)], predict(fit.dat), ylim=c(0,100))

summary(fit.dat)

**************************************************************************
Formula: dat ~ SSlogis(x, Asym, xmid, scal)

Parameters:
     Estimate Std. Error t value Pr(>|t|)
Asym   85.651      1.716  49.900  < 2e-16 ***
xmid   72.214      1.036  69.697  < 2e-16 ***
scal   -6.150      0.850  -7.236  7.9e-11 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.33 on 105 degrees of freedom

Number of iterations to convergence: 10
Achieved convergence tolerance: 4.405e-06
  (45 observations deleted due to missingness)
**************************************************************************

>From r-help, SSlogis parameters asym, xmid and scal are defined as:

Asym: a numeric parameter representing the asymptote.

xmid: a numeric parameter representing the x value at the inflection
point of the curve. The value of SSlogis will be Asym/2 at xmid.

scal: a numeric scale parameter on the input axis.

and it states that the value of SSlogis "is a numeric vector of the
same length as input. It is the value of the expression
sym/(1+exp((xmid-input)/scal)). If all of the arguments Asym, xmid,
and scal are names of objects the gradient matrix with respect to
these names is attached as an attribute named gradient."

However, how do I get the actual function for the curve that is
generated? I don't think it can just be: y=
asym/((1+e^((xmid-x)/scal)))?

Also, how do you determine the starting parameters to input in for
asym, xmin, and scal?

Perhaps I need to start at the beginning and define my own function,
and not rely on SSlogis to provide it?

What I want to be able to do is determine a local maximum for my curve
(the x value at which this curve inflects (the upper inflection)), and
the x value for the local minimum (the lower inflection curve), and
the x value counts in between these values. I think in order to do
this I need to differentiate the function.

Any insight on this would be greatly appreciated.

Sincerely,

Katrina